How many moles of S are needed to combine with 0.208 mol Al to give Al2S3?
break it down thanks
To determine the number of moles of S needed to combine with 0.208 mol of Al to produce Al2S3, we can refer to the balanced chemical equation for the reaction:
2 Al + 3 S → Al2S3
From the equation, we can see that it takes 3 moles of S to react with 2 moles of Al to produce 1 mole of Al2S3.
Now, we need to calculate the number of moles of S required. We know that the molar ratio of S to Al2S3 is 3 moles of S to 1 mole of Al2S3.
Given that we have 0.208 moles of Al, we can set up a proportion:
(3 moles of S / 2 moles of Al) = (x moles of S / 0.208 moles of Al)
Cross-multiplying this proportion, we get:
(3 moles of S) * (0.208 moles of Al) = (2 moles of Al) * (x moles of S)
0.624 moles of S = 2x
Now, let's solve for x by dividing both sides of the equation by 2:
0.624 moles of S / 2 = x
x = 0.312 moles of S
Therefore, it would require 0.312 moles of S to combine with 0.208 moles of Al to form Al2S3.
To determine the number of moles of S needed to combine with 0.208 mol Al to give Al2S3, we first need to analyze the balanced chemical equation for the formation of Al2S3.
The balanced equation is:
2Al + 3S → Al2S3
From the equation, we can see that 2 moles of Al react with 3 moles of S to form 1 mole of Al2S3.
Given that we have 0.208 mol Al, we can use the mole ratio from the balanced equation to find the number of moles of S needed.
Using the mole ratio, we have:
0.208 mol Al × (3 mol S / 2 mol Al) = 0.312 mol S
Therefore, 0.312 moles of S are needed to combine with 0.208 moles of Al to give Al2S3.
It looks like 3 moles of S for each two moles of Al.
so the answer is... (3/2)*.208